If you come from Stack Overflow, using atan2 might be a simpler solution for you.

Let , . If is the "oriented" angle from to (that is, rotating by gives ), then:

In Matlab, this is equivalent to wrapToPi(angle(x2+i*y2) - angle(x1+i*y1)).

 atan2(vector1.y - vector2.y, vector1.x - vector2.x)

is the angle between the difference vector (connecting vector2 and vector1) and the x-axis, which is problably not what you meant.

The (directed) angle from vector1 to vector2 can be computed as

angle = atan2(vector2.y, vector2.x) - atan2(vector1.y, vector1.x);

and you may want to normalize it to the range [0, 2 π):

if (angle < 0) { angle += 2 * M_PI; }

or to the range (-π, π]:

if (angle > M_PI)        { angle -= 2 * M_PI; }
else if (angle <= -M_PI) { angle += 2 * M_PI; }

[Not sure if this information is completely what you're looking for, but it certainly is relevant. Please give a more specific problem statement or a simple worked example and I'll be happy to expand/refine my answer.]

The angle between vectors $\vec{x}$ and $\vec{y}$ is defined using the dot product like so: $$ \cos(\theta) = \frac{\vec{x}\cdot \vec{y}}{\|\vec{x}\| \ \|\vec{y}\|}$$ where the expression $\|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ is the magnitude/norm of a vector. The magnitude of a vector in 3D space is just the square root of the sum of the squares of the $i,j,k$ components of that vector.

By using the inverse cosine function, you can determine the angle between the vectors. You'll have to pay attention to the sign of the dot product to determine if the resulting angle is acute (positive dot product), perpendicular (zero dot product), or obtuse (negative dot product).